Question

Find the energy at the bottom of the first allowed band, for the case$\mathbf{\beta }\mathbf{=}\mathbf{10}$ , correct to three significant digits. For the sake of argument, assume $\frac{\mathbf{\alpha }}{\mathbf{a}}\mathbf{=}\mathbf{1}$eV.

Short answer

The energy at the bottom of the first band is 0.345eV.

Step-by-step solution

Define the Schrödinger equation

• A differential equation that describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer is related to a particle's probability density in space and time.
• The time-dependent Schrodinger equation is represented as

$\mathit{I}\mathit{ħ}\frac{\mathbf{d}}{\mathbf{d}\mathbf{t}}\left|\psi \left(t\right)>=\hat{H}\right|\mathit{\psi }\left(t\right)\mathbf{>}$

Calculating the minimum energy of first band

The minimum energy of the first band required z,f(z)=1

And f(z) is given by

$$\begin{gathered} f\left(z\right)=\cos z+\mathrm{\beta }\frac{\sin \mathrm{z}}{\mathrm{z}} \\ \mathrm{z}=\mathrm{ka},\mathrm{z}\le \mathrm{\pi \beta }=10=\frac{\mathrm{m\alpha a}}{{\mathrm{h}}^2} \\ \Rightarrow \mathrm{f}\left(\mathrm{z}\right)=\cos \mathrm{z}+10\frac{\sin \mathrm{z}}{\mathrm{z}} \end{gathered}$$

Using MATLAB to calculate the minimum energy of first band

$$\begin{gathered} \mathrm{Using}\mathrm{MATLAB}\mathrm{we}\mathrm{get}\mathrm{z}=2.6276.\mathrm{Energy}\mathrm{is}\mathrm{given}\mathrm{with} \\ \mathrm{E}=\frac{{\mathrm{h}}^2{\mathrm{k}}^2}{2\mathrm{m}} \\ =\frac{{\mathrm{h}}^2}{2\mathrm{m}}{\left(\frac{\mathrm{z}}{\mathrm{a}}\right)}^2=\frac{{\mathrm{z}}^2}{2\mathrm{a}}\frac{{\mathrm{h}}^2}{\mathrm{m\beta }}\frac{\mathrm{\alpha }}{\mathrm{\alpha }} \\ =\frac{{\mathrm{z}}^2}{2\mathrm{a}}\frac{\mathrm{\alpha }}{\mathrm{\beta }} \\ \frac{\mathrm{\alpha }}{\mathrm{a}}=1\mathrm{eV} \\ \mathrm{E}=\frac{2.{62767}^2}{20}1\mathrm{eV} \\ \mathrm{E}=0.345\mathrm{eV} \end{gathered}$$

Therefore the energy at the bottom of the first band is 0.345eV.